Switching — circuit, packet, virtual circuit: Exercises

This page is a self-testing ladder for Switching — circuit, packet, virtual circuit. Work each problem before opening its solution. Every symbol used here was defined in the parent note; a quick reminder table sits below so you never have to guess.

Recall The symbols we reuse (open only if you forgot)

$L$ ::: message size in bits
$R$ ::: link rate in bits per second (bps)
$d$ ::: number of links (hops) on the path
$t_{p}$ ::: propagation delay per link (seconds)
$S$ ::: circuit setup time (seconds)
$P$ ::: number of packets we split $L$ into
$\frac{L}{R}$ ::: transmission time — how long to push all bits onto one link

Level 1 — Recognition

Exercise 1.1

For each network below, name the switching type: (a) a landline phone call reserving a path end-to-end, (b) web packets each carrying a full IP destination address forwarded independently, (c) MPLS forwarding by swapping small labels along a pre-set path.

Recall Solution

WHAT we do: match each description to the defining property.

(a) A reserved end-to-end path used exclusively → circuit switching.
(b) Each unit carries a full destination address and is forwarded independently → packet (datagram) switching.
(c) A pre-set path plus label swapping with no bandwidth reserved → virtual circuit switching.

WHY: the three types differ by when the path is chosen and what each packet carries. "Full address per packet" is the datagram fingerprint; "small label + fixed path" is the VC fingerprint.

Exercise 1.2

True or false, with one-line reason each: (a) Circuit switching keeps per-connection state in switches. (b) Datagram switching guarantees in-order delivery. (c) A virtual circuit reserves the full link bandwidth.

Recall Solution

(a) True. A reserved circuit is per-connection state (the input↔output link binding).
(b) False. Datagram packets may take different paths and arrive out of order; ordering is the job of a higher layer like TCP.
(c) False. A VC reserves a logical path only; links are still shared, bandwidth is not dedicated.

Level 2 — Application

Exercise 2.1

A single packet of $L = 2 \times 1 0^{6}$ bits crosses $d = 4$ links at rate $R = 1 0^{6}$ bps with propagation $t_{p} = 2$ ms per link, using datagram store-and-forward. Find the total delay.

Recall Solution

WHAT / WHY: one packet, store-and-forward ⇒ each of the $4$ links must fully receive then resend, so transmission time appears once per hop. $\frac{L}{R} = \frac{2 \times 1 0 ^{6}}{1 0 ^{6}} = 2 s, d t_{p} = 4 (0.002) = 0.008 s$ $T = d \cdot \frac{L}{R} + d t_{p} = 4 (2) + 0.008 = 8.008 s$

Exercise 2.2

Same message and path as 2.1, but sent as a circuit with setup $S = 50$ ms. Find the delay and explain the difference from 2.1.

Recall Solution

WHAT / WHY: after setup the path is a continuous pipe — no store-and-forward — so $\frac{L}{R}$ appears only once. $T = S + d t_{p} + \frac{L}{R} = 0.05 + 0.008 + 2 = 2.058 s$ Difference: datagram cost $8.008$ s because each hop replayed all $2$ Mbit serially ( $4 \times 2$ s). Circuit paid a one-time $50$ ms setup but transmits once, so it is nearly 4× faster for this single big packet.

Level 3 — Analysis

Exercise 3.1

Take $L = 1 0^{6}$ bits, $R = 1 0^{6}$ bps, $d = 5$ hops, $t_{p} = 1$ ms. Compute datagram delay for $P = 1$ packet and for $P = 100$ packets (ignore headers). By how much does pipelining cut the delay?

Recall Solution

Reference the pipeline picture:

Figure — Switching — circuit, packet, virtual circuit

WHAT / WHY: splitting into

P

packets lets link 2 carry packet 1 while link 1 already carries packet 2 — parallel use of links.

\frac{L}{R} = 1 s, d t_{p} = 5 (0.001) = 0.005 s

P = 1

T = \frac{5 + 1 - 1}{1} \cdot 1 + 0.005 = 5 + 0.005 = 5.005 s

P = 100

T = \frac{5 + 100 - 1}{100} \cdot 1 + 0.005 = 1.04 + 0.005 = 1.045 s

Saving:

5.005 - 1.045 = 3.96

s — roughly 5× faster, because

\frac{d + P - 1}{P} = \frac{104}{100} \to 1

P

grows, driving transmission cost toward a single

L / R

Exercise 3.2

For the numbers in 3.1, what value of $P$ makes the effective transmission factor $\frac{d + P - 1}{P}$ equal to exactly $1.1$ ? What does this tell you about diminishing returns?

Recall Solution

WHAT: solve $\frac{d + P - 1}{P} = 1.1$ with $d = 5$ . $5 + P - 1 = 1.1 P ⟹ 4 = 0.1 P ⟹ P = 40$ WHY it matters: at $P = 40$ we are already within $10%$ of the ideal single-hop transmission. Doubling $P$ again to $80$ only pushes the factor from $1.1$ to $1.05$ — a tiny gain. The curve flattens, and in reality every packet also carries a header, so past a point more packets = more header overhead = slower. Pipelining has a sweet spot, not "infinitely small is best."

Level 4 — Synthesis

Exercise 4.1

Compare all three schemes for one file: $L = 4 \times 1 0^{6}$ bits, $R = 1 0^{6}$ bps, $d = 3$ hops, $t_{p} = 1$ ms, circuit setup $S = 80$ ms. Split into $P = 500$ packets for the pipelined datagram case. Rank them fastest→slowest and explain the ranking.

Recall Solution

$\frac{L}{R} = 4 s$ , $d t_{p} = 0.003 s$ .

Circuit: $T = 0.08 + 0.003 + 4 = 4.083 s$
Datagram, 1 packet: $T = 3 (4) + 0.003 = 12.003 s$
Datagram, 500 packets: $T = \frac{3 + 500 - 1}{500} \cdot 4 + 0.003 = \frac{502}{500} (4) + 0.003 = 4.016 + 0.003 = 4.019 s$

Ranking (fastest first): pipelined datagram ( $4.019$ ) < circuit ( $4.083$ ) < single-packet datagram ( $12.003$ ). WHY: pipelined datagram gets transmission down to ≈ $L / R$ and pays no $80$ ms setup, so it edges out circuit. Single-packet datagram is crippled by store-and-forward replaying all $4$ Mbit on each of $3$ hops. This is precisely why the modern internet is packet-switched — see how it beats the reserved circuit.

Exercise 4.2

Along a virtual circuit A→B, the VC tables are:

Switch S1: (in port 1, VC 7) → (out port 2, VC 3)
Switch S2: (in port 5, VC 3) → (out port 4, VC 9)

A sends a packet labeled 7. Trace the label at each hop and state what B ultimately receives. Then explain why S1 and S2 can both freely reuse small numbers.

Recall Solution

Trace (label swapping):

A → S1 with label 7. S1 looks up (port 1, VC 7) → outputs on port 2 with label 3.
S1 → S2 with label 3. S2 looks up (port 5, VC 3) → outputs on port 4 with label 9.
S2 → B with label 9.

WHY reuse works: each label is locally meaningful only — VC 3 at S1's output means nothing to any other switch. A globally unique number would force coordination across the whole network; local swapping lets every switch manage its own small label space independently. This is the trick behind ATM and MPLS.

Level 5 — Mastery

Exercise 5.1

A datagram network sends $L = 1 0^{6}$ bits over $d = 4$ hops, $R = 1 0^{6}$ bps, $t_{p} = 0.5$ ms, but each packet now carries a header of $h = 200$ bits. If we split into $P$ packets, the total bits transmitted per hop are $L + P h$ and each packet has size $\frac{L}{P} + h$ . The pipelined delay becomes $T (P) = \frac{( d - 1 ) ( \frac{L}{P} + h ) + P ( \frac{L}{P} + h )}{R} + d t_{p} = \frac{( d - 1 + P ) ( \frac{L}{P} + h )}{R} + d t_{p} .$ Compute $T$ for $P = 100$ and $P = 2000$ , and say which is better — proving the L3 trap is real.

Recall Solution

Reference the overhead curve:

d t_{p} = 4 (0.0005) = 0.002 s

$P = 100$ : packet size $= \frac{1 0 ^{6}}{100} + 200 = 10, 200$ bits. $T = \frac{( 4 - 1 + 100 ) ( 10 , 200 )}{1 0 ^{6}} + 0.002 = \frac{103 \times 10 , 200}{1 0 ^{6}} + 0.002 = \frac{1 , 050 , 600}{1 0 ^{6}} + 0.002 = 1.0526 s$

$P = 2000$ : packet size $= \frac{1 0 ^{6}}{2000} + 200 = 700$ bits. $T = \frac{( 4 - 1 + 2000 ) ( 700 )}{1 0 ^{6}} + 0.002 = \frac{2003 \times 700}{1 0 ^{6}} + 0.002 = \frac{1 , 402 , 100}{1 0 ^{6}} + 0.002 = 1.4041 s$

Verdict: $P = 100$ ( $\approx 1.053$ s) beats $P = 2000$ ( $\approx 1.404$ s). With headers present, splitting too finely adds $P h = 400, 000$ header bits, which more than cancels the pipelining gain. This proves the L3 warning: there is a finite optimal $P$ , not "smaller is always better."

Exercise 5.2

Design decision: a provider carries bursty web traffic (short bursts, long idle gaps) and steady video calls (constant bitrate, needs strict ordering, low overhead). Recommend a switching type for each and justify using the trade-off table from the parent note.

Recall Solution

Bursty web → datagram packet switching. Bursty means links sit idle most of the time; reserving bandwidth (circuit / dedicated slots) would waste it. Datagram statistically multiplexes many bursty users onto shared links — no reservation, no idle waste. Out-of-order arrival is fine because TCP reorders. See Queueing Delay for the cost this shifts onto buffers. Steady video → virtual circuit. A call is constant-rate and needs ordered, low-overhead delivery. A VC gives a fixed path (ordering guaranteed), tiny labels (low header), and cheap label-swap forwarding — without wasting bandwidth on idle gaps (there aren't many). Pure circuit switching would also work but wastes bandwidth on any pause and needs full reservation; VC is the modern compromise (ATM / MPLS were built for exactly this). Route choice at setup is handled by Routing Algorithms.

Recall Quick self-check (cover the answers)

Circuit delay formula ::: $S + d t_{p} + \frac{L}{R}$ Why no $\times d$ on transmission in circuit switching ::: no store-and-forward — the path is a continuous pipe after setup Pipelined datagram transmission factor as $P \to \infty$ ::: $\frac{d + P - 1}{P} \to 1$ What limits how far we shrink packets ::: per-packet header overhead $P h$ What a VC label does at each switch ::: it is looked up and swapped (local lookup key, not an address)